LeetCode 1199: Minimum Time to Build Blocks Solution

Master LeetCode problem 1199 (Minimum Time to Build Blocks), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1199. Minimum Time to Build Blocks

Hard

Array Math Greedy Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We can iterate over all possible split points and calculate the minimum time required to build blocks for each split point. The key insight is that the minimum time to build blocks at a specific split point is the maximum time needed to build the left block and the right block, plus the time required to build the current block.

Sort the blocks in non-decreasing order.
Initialize a 2D array dp of size n+1 by n+1 where n is the number of blocks. dp[i][j] will represent the minimum time required to build j blocks from the first i blocks.
Initialize the base cases: dp[1][1] = blocks[0] and dp[2][1] = max(blocks[0], blocks[1]).
Iterate over all possible split points k from 2 to n and update dp[i][j] using the formula dp[i][j] = min(dp[i][j], max(dp[k][j-1], prefix_sum[i] - prefix_sum[k])).
The final answer will be dp[n][split] where split is the number of blocks to build in parallel.

Time complexity: O(n^3)
Space complexity: O(n^2)

Solution Code

class Solution {
    public int minBuildTime(int[] blocks, int split) {
        Arrays.sort(blocks);
        int n = blocks.length;
        int[][] dp = new int[n + 1][n + 1];
        int[] prefixSum = new int[n + 1];
        
        for (int i = 1; i <= n; i++) {
            prefixSum[i] = prefixSum[i - 1] + blocks[i - 1];
        }
        
        for (int i = 1; i <= n; i++) {
            dp[i][1] = blocks[i - 1];
        }
        
        for (int i = 2; i <= n; i++) {
            dp[i][1] = Math.max(dp[i][1], blocks[i - 1]);
        }
        
        for (int j = 2; j <= n; j++) {
            for (int i = j; i <= n; i++) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = 2; k <= i; k++) {
                    dp[i][j] = Math.min(dp[i][j], Math.max(dp[k][j - 1], prefixSum[i] - prefixSum[k]));
                }
            }
        }
        
        return dp[n][split];
    }
}

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